Manufacture of a certain component requires three different machining operations. The amount of time each operation...
Manufacture of a certain component requires three different machining operations. The amount of time each operation requires (operation time) is normally distributed with mean 10 and variance 4. The three operation times are independent.
Referring to the previous manufacturing example, now suppose that the cost for the first machining operation is $1 per minute. That for the second and third operations are $2 per minute and $3 per minute, respectively. What is the probability that total cost for making the next component is more than $70?
thanks
Homework Answers
here total cost X =X1+2X2+3X3
tehrefore mean of X =10+2*10+3*10=60
and std deviation =sqrt(4+22*4+32*4)=7.483
probability that total cost for making the next component is more than $70 =P(X>70)=P(Z>(70-60)/7.483)=P(Z>1.34)
=0.0901
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